(C) Given $f(x) = (x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}$ for $x \neq a$ and $f(a) = 0$. To check continuity at $x=a$,we evaluate the l

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The function $f(x)$ is continuous at $x=a$

(C) Given $f(x) = (x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}$ for $x \neq a$ and $f(a) = 0$. To check continuity at $x=a$,we evaluate the limits: Left-hand limit: $\lim_{x \rightarrow a^-} f(x) = \lim_{h \rightarrow 0} (-h) \frac{e^{-1/h}-1}{e^{-1/h}+1} = 0 \times \frac{0-1}{0+1} = 0$. Right-hand limit: $\lim_{x \rightarrow a^+} f(x) = \lim_{h \rightarrow 0} (h) \frac{e^{1/h}-1}{e^{1/h}+1} = \lim_{h \rightarrow 0} h \frac{1-e^{-1/h}}{1+e^{-1/h}} = 0 \times \frac{1-0}{1+0} = 0$. Since $\lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x) = f(a) = 0$,the function $f(x)$ is continuous at $x=a$.

Class 12 Mathematics Continuity and Differentiation Continuity

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Define $f: R \rightarrow R$ by $f(x) = \begin{cases} (x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}, & x \neq a \\ 0, & x=a \end{cases}$. Then which one of the following is true?

TS EAMCET 2020 All TS EAMCET

A
Left and right limits of $f$ at $x=a$ are equal and they are not equal to $f(a)$
B
Both left and right limits of $f$ at $x=a$ exist and are not equal
C
The function $f(x)$ is continuous at $x=a$
D
The function $f(x)$ has a simple discontinuity at a point other than $a$

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Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a function defined by $f(x) = [\frac{x}{2} + 3] - [\sqrt{x}]$. Let $S$ be the set of all points in the interval $[0, 8]$ at which $f$ is not continuous. Then $\sum_{a \in S} a$ is equal to:

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Let $a, b, c$ be three real numbers. If the function $f(x) = \begin{cases} \cos(2x + \pi) & \text{if } x \leq 0 \\ ax^2 + b & \text{if } 0 < x < 1 \\ cx + 4 & \text{if } 1 \leq x \leq 2 \\ 3a + 1 & \text{if } x \geq 2 \end{cases}$ is continuous everywhere,then $b^2 - bc + c^2 =$

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If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{1+3x^2-\cos 2x}{x^2}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then $k$ is equal to

MediumAP EAMCET 2010

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Let $f(x) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\sin x} \right)}^{2n}}}}{{{3^n} - {{\left( {2\cos x} \right)}^{2n}}}}; n \in Z$,$x \ne m\pi \pm \frac{\pi }{6}; m \in Z$ and $f\left( {m\pi \pm \frac{\pi }{6}} \right) = 0$. Then which of the following is true?

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Let $[t]$ represent the greatest integer not exceeding $t$. Then the number of points of discontinuity of $f(x) = [10^x]$ in the interval $(0, 10)$ is:

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Define $f: R \rightarrow R$ by $f(x) = \begin{cases} (x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}, & x \neq a \\ 0, & x=a \end{cases}$. Then which one of the following is true?

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Let $a, b, c$ be three real numbers. If the function $f(x) = \begin{cases} \cos(2x + \pi) & \text{if } x \leq 0 \\ ax^2 + b & \text{if } 0 < x < 1 \\ cx + 4 & \text{if } 1 \leq x \leq 2 \\ 3a + 1 & \text{if } x \geq 2 \end{cases}$ is continuous everywhere,then $b^2 - bc + c^2 =$

If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{1+3x^2-\cos 2x}{x^2}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then $k$ is equal to

Let $f(x) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\sin x} \right)}^{2n}}}}{{{3^n} - {{\left( {2\cos x} \right)}^{2n}}}}; n \in Z$,$x \ne m\pi \pm \frac{\pi }{6}; m \in Z$ and $f\left( {m\pi \pm \frac{\pi }{6}} \right) = 0$. Then which of the following is true?

Let $[t]$ represent the greatest integer not exceeding $t$. Then the number of points of discontinuity of $f(x) = [10^x]$ in the interval $(0, 10)$ is:

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